$xy$-symmetry of the electric potential: Charged sphere at an external uniform electric field

Griffiths Introduction to Electrodynamics problem 3.21

Find the potential outside a charged metal sphere of charge $$Q$$ and radius $$R$$, placed in an otherwise uniform electric field $$E_0\hat{z}$$.

The Instructor's solution manual says that the electric potential has inverse $$xy$$-plane symmetry, so it can be solved as the same way as the uncharged sphere case, but I can't see why this symmetry holds.

In the uncharged sphere problem, inverse $$xy$$-symmetry holds since the upper/lower hemispheres have the same amount of charge, therefore having the same charge distribution. However, if the net charge is, say $$Q>0$$, positive charges will be concentrated at the upper hemisphere since there is fewer negative charges to resist to this movement. So I think the symmetry doesn't hold anymore. Is there anything wrong here?

//EDIT: inverse $$xy$$-symmetry=inverse symmetry of the $$xy$$-plane, $$\rho$$ at $$(x,y,z)$$ and $$-\rho$$ at $$(x,y,-z)$$.

• One might indeed say that the charge density $\rho(z)$ is antisymmetric under reflection about the x,y plane. Note how the rotational symmetry of the problem eliminates the x,y dependence of $\rho$. Feb 27 at 0:35

The conducting sphere has an induced charge distribution with no net charge, and the external electric field is uniform and in the $$\hat z$$ direction. So the resultant field is such that it looks like the uniform case when far from the sphere, and when near it, turns in to meet the conducting surface in a way everywhere parallel to the surface normal. This problem has rotational symmetry about the z-axis. For example, it doesn't matter at which axis the x or y along which you approach the equator, the charge density looks the same in each case; this kind of symmetry holds for any slice parallel to the x,y plane; so indeed there is x,y symmetry (rotation about z-axis).

So, I looked up problem 3.8, and when Griffiths talks about using symmetry to set the whole x,y plane to zero, he is referring to the asymmetry of the charge density: $$\rho(z)=-\rho(-z)$$.

• No, the conducting sphere is not uniformly charged. The charges in the surface of the metal is attracted by the electric field, forming an induced dipole.
– Leo
Feb 26 at 0:06
• @Leo Yes you are correct, the uniform electric field disperses the charge in the conductor; I think I was thinking about the electric field being uniform.. Second blunder I have made today; my allergies must be getting the best of me! I'll edit that off, everything else in the post holds though. Feb 26 at 2:29
• Sorry, but I didn't mean rotation wrt z axis when I said $xy$-symmetry. I meant symmetry of the $xy$-plane, $(x,y,z)$ and $(x,y,-z)$.
– Leo
Feb 27 at 0:17
• @Leo Well, it is clear, as you know yourself, that there is no way the sphere will have a symmetric charge density when reflected across the x,y plane. If that is what the solution manual meant and not x,y independence, then it is certainly erroneous. Feb 27 at 0:22
• When the net charge is 0, it has $xy$-symmetry. When there is charge density $\rho$ at $(x,y,z)$, there is $-\rho$ at $(x,y,-z)$.
– Leo
Feb 27 at 0:27